How far should an object be placed from a convex lens of focal length 20cm to obtain its image at a distance of 30cm from the lens? What will be the height of the image, if the objects is 6cm tall?
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According to the question; Object distance = u; Image distance (v) = 30cm; Focal length = 20cm By lens formula;$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{30}-\frac{1}{\text{u}}=\frac{1}{20}$
$\Rightarrow\frac{1}{\text{u}}=\frac{1}{30}-\frac{1}{20}$
$\Rightarrow\frac{1}{\text{u}}=\frac{20-30}{600}=\frac{-10}{600}$
$\Rightarrow\frac{1}{\text{u}}=\frac{-1}{60}$
$\Rightarrow\text{u}=60\text{cm.}$
Therefore object is placed at 60cm in front of lens. Now; Height of object $h_1=6 cm$; Magnification $=\frac{\text{H}_2}{\text{H}_1}=\frac{\text{v}}{\text{u}}$ Putting values of v and u Magnification $=\frac{\text{H}_2}{6}=\frac{30}{-60}$$\Rightarrow\frac{\text{h}_2}{6}=-\frac{1}{2}$
$\Rightarrow\text{h}_2=-\frac{6}{2}=-3$
Height of image is 3cm. Negative sign means image is real and inverted.
$\Rightarrow\frac{1}{30}-\frac{1}{\text{u}}=\frac{1}{20}$
$\Rightarrow\frac{1}{\text{u}}=\frac{1}{30}-\frac{1}{20}$
$\Rightarrow\frac{1}{\text{u}}=\frac{20-30}{600}=\frac{-10}{600}$
$\Rightarrow\frac{1}{\text{u}}=\frac{-1}{60}$
$\Rightarrow\text{u}=60\text{cm.}$
Therefore object is placed at 60cm in front of lens. Now; Height of object $h_1=6 cm$; Magnification $=\frac{\text{H}_2}{\text{H}_1}=\frac{\text{v}}{\text{u}}$ Putting values of v and u Magnification $=\frac{\text{H}_2}{6}=\frac{30}{-60}$$\Rightarrow\frac{\text{h}_2}{6}=-\frac{1}{2}$
$\Rightarrow\text{h}_2=-\frac{6}{2}=-3$
Height of image is 3cm. Negative sign means image is real and inverted.
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